Given :
p =$7000,
A = $10000,
r = 4.3% = 0.043
Assume a simple interest
[tex]\begin{gathered} A=P+I \\ I=P\cdot r\cdot t \\ A=P+P\cdot r\cdot t \\ A=P\cdot(1+r\cdot t) \end{gathered}[/tex]So,
[tex]\begin{gathered} A=P\cdot(1+r\cdot t) \\ 10000=7000\cdot(1+0.043\cdot t) \end{gathered}[/tex]Solve for t :
[tex]\begin{gathered} 10000=7000\cdot(1+0.043\cdot t) \\ 1+0.043\cdot t=\frac{10000}{7000}=\frac{10}{7} \\ 0.043\cdot t=\frac{10}{7}-1 \\ \\ 0.043t=0.42857 \\ \\ t=\frac{0.42857}{0.043}\approx9.97 \end{gathered}[/tex]So, rounding to the nearest year
So, time = 10 years